This is a textbook exercise that might be seen as a generalisation of Pearson's product-moment correlation coefficient between two variables, with multiple groups each having variable sizes:

Let there be $k$ groups of data of size $n_j$ on two variables $(x,y)$, with means $(\bar x_j,\bar y_j)$, variances $(s_{xj}^2,s_{yj}^2)$, and correlation coefficient $r_j$; $j=1,2,\ldots,k$. Then the correlation coefficient of the combined data of size $\sum_{j=1}^kn_j$ is given by

$$r=\frac{\sum_{j=1}^kn_jr_js_{xj}s_{yj}+\sum_{j=1}^k n_j(x_j-\bar x)(y_j-\bar y)}{\sqrt{\sum_{j=1}^kn_js_{xj}^2+\sum_{j=1}^kn_j(\bar x_j-\bar x)}\sqrt{\sum_{j=1}^kn_js_{yj}^2+\sum_{j=1}^kn_j(\bar y_j-\bar y)}}\quad,$$

where $\bar x$ and $\bar y$ are the grand means of $x$ and $y$ respectively.

The exercise also asks to explain from the above formula that $r_j$ may be zero for each $j$ and yet $r$ may be non-zero

While I am not that interested in an analytical derivation of the above expression, I would like to have an intuitive understanding of the fact that the individual correlations may vanish and still the overall correlation may be non-zero. But I am not looking for this verification from the formula. Is there a visual explanation that one can come up with?

By the way, is there a particular name for $r$ as defined above? Any reference where this shows up in descriptive statistics will be great.

  • 1
    $\begingroup$ The $r$ defined above is just the ordinary Pearson correlation computed from $x$ and $y$ across all the groups. So it isn't a generalisation and it doesn't have a special name. $\endgroup$ Aug 5, 2018 at 10:26

1 Answer 1


This is one option:

There is zero correlation for groups within vertical lines. But overall correlation is positive.

correlation scatter

  • $\begingroup$ Yea, you are right. Editing to a better example. $\endgroup$
    – yoav_aaa
    Aug 5, 2018 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.